Solve the following system of equations:
$2x\ –\ \left(\frac{3}{y}\right)\ =\ 9$
$3x\ +\ \left(\frac{7}{y}\right)\ =\ 2,\ y\ ≠\ 0$

Given:

The given system of equations is:

$2x\ –\ \left(\frac{3}{y}\right)\ =\ 9$

$3x\ +\ \left(\frac{7}{y}\right)\ =\ 2,\ y\ ≠\ 0$

To do:

We have to solve the given system of equations.

Solution:

The given system of equations can be written as,

$2x-\frac{3}{y}=9$

Let $\frac{1}{y}=k$,

$\Rightarrow 2x-3k=9$---(i)

$3x+\frac{7}{y}=2$

$\Rightarrow 3x+7k=2$

$\Rightarrow 7k=2-3x$

$\Rightarrow k=\frac{2-3x}{7}$----(ii)

Substitute $k=\frac{2-3x}{7}$ in equation (i), we get,

$2x-3(\frac{2-3x}{7})=9$

$2x-\frac{3(2-3x)}{7}=9$ 

Multiplying by $7$ on both sides, we get,

$7(2x)-7(\frac{6-9x}{7})=7(9)$

$14x-(6-9x)=63$

$14x-6+9x=63$

$23x=63+6$

$23x=69$

$x=\frac{69}{23}$

$x=3$

Substituting the value of $x=3$ in equation (ii), we get,

$k=\frac{2-3(3)}{7}$

$k=\frac{2-9}{7}$

$k=\frac{-7}{7}$

$k=-1$

This implies,

$y=\frac{1}{k}=\frac{1}{-1}$

$y=-1$

Therefore, the solution of the given system of equations is $x=3$ and $y=-1$.

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Updated on: 10-Oct-2022

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